A spherical ball of lead $3\ cm$ in diameter is melted and recast into three spherical balls. It the diameters of two balls be $\frac{3}{2}\ cm$ and $2\ cm$, find the diameter of the third ball.

 Given:

A spherical ball of lead $3\ cm$ in diameter is melted and recast into three spherical balls.

The diameters of two balls are $\frac{3}{2}\ cm$ and $2\ cm$.

To do:

We have to find the diameter of the third ball.

Solution:

Diameter of the spherical ball of lead $= 3\ cm$

This implies,

Radius of the spherical ball of lead $(r)=\frac{3}{2} \mathrm{~cm}$

Therefore,

Volume of the ball of lead $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}$

$=\frac{9 \pi}{2} \mathrm{~cm}^{3}$

Diameter of the 1st smaller ball $=\frac{3}{2} \mathrm{~cm}$

This implies,

Radius of the 1st smaller ball $(r_{1})=\frac{3}{4} \mathrm{~cm}$

Volume of the 1st smaller ball $=\frac{4}{3} \pi(r_{1})^{3}$

$=\frac{4}{3} \pi \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}$

$=\frac{9}{16} \pi \mathrm{cm}^{3}$

Diameter of the second smaller ball $=2 \mathrm{~cm}$

This implies,

Radius of the second smaller ball $(r_{2})=\frac{2}{2}$

$=1 \mathrm{~cm}$

Therefore,

Volume of the second ball $=\frac{4}{3} \pi(r_{2})^{3}$

$=\frac{4}{3} \pi \times 1^3$

$=\frac{4}{3} \pi \mathrm{cm}^{3}$

Volume of the third smaller ball $=\frac{9}{2} \pi-(\frac{9}{16} \pi+\frac{4}{3} \pi)$

$=(\frac{9}{2}-\frac{9}{16}-\frac{4}{3}) \pi$

$=\frac{216-27-64}{48}$

$=\frac{125}{48} \pi \mathrm{cm}^{3}$

Therefore,

Radius of the third ball $=(\frac{\text { Volume } \times 3}{4 \pi})^{\frac{1}{3}}$

$=(\frac{125 \pi \times 3}{48 \times 4 \pi})^{\frac{1}{3}}$

$=(\frac{125}{64})^{\frac{1}{3}}$

$=[(\frac{5}{4})^{3}]^{\frac{1}{3}}$

$=\frac{5}{4}$

$=1.25 \mathrm{~cm}$

Diameter of third ball $=2 \times$ radius

$=2 \times 1.25$

$=2.5 \mathrm{~cm}$

Tutorialspoint

Updated on: 10-Oct-2022

46 Views

Kickstart Your Career

Get certified by completing the course

Get Started